Radius of Convergence from Limit of Sequence/Complex Case

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Let $\xi \in \C$ be a complex number.

Let $\displaystyle S \paren z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ be a (complex) power series about $\xi$.

Let the sequence $\sequence {\cmod {\dfrac {a_{n + 1} } {a_n} } }_{n \mathop \in \N}$ converges.

Then $R$ is given by:

$\displaystyle \dfrac 1 R = \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} }$


$\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} } = 0$

then the radius of convergence is infinite, and $S \paren z$ is absolutely convergent for all $z \in \C$.


Let the sequence $\sequence {\cmod {\dfrac {a_{n+1} } {a_n} } }_{n \mathop \in \N}$ converge.

Let $\epsilon \in \R_{>0}$, and let $z \in \C$.

Let $\cmod {z - \xi} = R - \epsilon$.

By definition of radius of convergence, it follows that $S \paren z$ is absolutely convergent.

Then from the Ratio Test:

$\lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n+1} \paren {z - \xi}^{n + 1} } {a_n \paren {z - \xi}^n} } \le 1$

By Multiple Rule for Complex Sequences, this inequality can be rearranged to obtain:

\(\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} }\) \(\le\) \(\displaystyle \cmod {\dfrac {\paren {z - \xi}^{n + 1} } {\paren {z - \xi}^n } }^{-1}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {\cmod {x - \xi}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {R - \epsilon}\)


Let $\cmod {z - \xi} = R + \epsilon$.

Then $S \paren z$ is divergent, so the Ratio Test shows that:

$\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} \paren {z - \xi}^{n + 1} } {a_n \paren {z - \xi}^n} } \ge 1$

Similarly, this inequality can be rearranged as:

$\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} } \ge \dfrac 1 {R + \epsilon}$


As $\epsilon > 0$ was arbitrary, it follows that:

$\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1}} {a_n} } = \dfrac 1 R$

Hence the result.


Also presented as

This result can also be seen presented as:

$\displaystyle R = \lim_{n \mathop \to \infty} \cmod {\frac {a_{n - 1} } {a_n} }$

but in this case the condition under which the radius of convergence is infinite is less conveniently stated.